Solving Systems Of Equations Finding X And Z Values
Hey guys! Today, we're diving into the exciting world of solving systems of equations. Specifically, we're going to tackle a problem where we need to find the values of x and z that satisfy two given equations. This is a fundamental concept in algebra, and mastering it will open doors to more advanced mathematical topics. So, let's get started and break down how to solve this type of problem step-by-step.
Understanding Systems of Equations
Before we jump into the solution, let's make sure we're all on the same page about what a system of equations actually is. A system of equations is simply a set of two or more equations that share the same variables. Our goal is to find values for these variables that make all the equations in the system true simultaneously. Think of it like finding a common ground where both equations agree. In our case, we have two equations with two variables, x and z. This is a classic setup, and there are several methods we can use to solve it.
The beauty of systems of equations lies in their ability to model real-world scenarios. From calculating the optimal mix of ingredients in a recipe to predicting the intersection point of two moving objects, these systems pop up everywhere. By understanding how to solve them, you're not just learning a math skill; you're gaining a powerful tool for problem-solving in various fields. So, pay close attention, and let's unravel the mystery of these equations together!
There are primarily two main methods that we can use to solve such equations. They are the substitution method and the elimination method. We will focus on the elimination method in this article.
The Elimination Method
Alright, let's get our hands dirty with the elimination method. This technique is super handy when you can manipulate the equations to cancel out one of the variables. The basic idea is to multiply one or both equations by a constant so that the coefficients of either x or z become opposites. When you add the equations together, that variable disappears, leaving you with a single equation in one variable. Sounds neat, right? Let's see how it works in practice with our example equations:
- 4x + 6z = 2
- 3x - 2z = 1
Looking at these equations, we can see that the coefficients of z are 6 and -2. A-ha! If we multiply the second equation by 3, the coefficient of z will become -6, which is the opposite of 6 in the first equation. This is exactly what we want! So, let's do that:
3 * (3x - 2z) = 3 * 1
This simplifies to:
- 9x - 6z = 3
Now, we have a new equation (equation 3) that's equivalent to equation 2, but with a z coefficient that's ready for elimination. The next step is where the magic happens: we add equation 1 and equation 3 together. This is a valid operation because we're essentially adding equal quantities to both sides of the overall system.
Step-by-Step Solution
Okay, let's walk through the solution step-by-step, making sure we don't miss anything. Remember, our goal is to find the values of x and z that satisfy both equations:
- 4x + 6z = 2
- 3x - 2z = 1
As we discussed, the first move is to multiply the second equation by 3. This will give us a -6z term, which is perfect for eliminating the z variable when we add the equations together. So, let's do it:
3 * (3x - 2z) = 3 * 1
This simplifies to:
- 9x - 6z = 3
Now, we're ready to add equation 1 and equation 3. Get your addition hats on, guys!
(4x + 6z) + (9x - 6z) = 2 + 3
Combine like terms:
13x = 5
Divide both sides by 13 to isolate x:
x = 5 / 13
Alright! We've found the value of x. That wasn't so bad, was it? Now, the next step is to plug this value of x back into one of the original equations to solve for z. You can choose either equation 1 or equation 2; it doesn't matter which one. Let's go with equation 1, just because.
4 * (5 / 13) + 6z = 2
Simplify:
20 / 13 + 6z = 2
Subtract 20/13 from both sides:
6z = 2 - 20 / 13
Find a common denominator and simplify:
6z = 26 / 13 - 20 / 13
6z = 6 / 13
Divide both sides by 6:
z = (6 / 13) / 6
z = 1 / 13
Boom! We've found the value of z as well. So, our solution is x = 5/13 and z = 1/13. That's how you crack a system of equations using the elimination method. We systematically eliminated one variable, solved for the other, and then plugged the result back in to find the remaining variable. Pretty slick, huh?
Checking the Solution
Before we declare victory, it's always a good idea to check our solution. This is a crucial step to make sure we haven't made any silly mistakes along the way. To check, we simply plug our values for x and z back into both of the original equations. If both equations hold true, then we know we've got the right answer. Let's do it!
First, let's check equation 1:
4x + 6z = 2
Plug in x = 5/13 and z = 1/13:
4 * (5 / 13) + 6 * (1 / 13) = 2
Simplify:
20 / 13 + 6 / 13 = 2
26 / 13 = 2
2 = 2
Great! Equation 1 checks out. Now, let's check equation 2:
3x - 2z = 1
Plug in x = 5/13 and z = 1/13:
3 * (5 / 13) - 2 * (1 / 13) = 1
Simplify:
15 / 13 - 2 / 13 = 1
13 / 13 = 1
1 = 1
Awesome! Equation 2 also holds true. Since our solution satisfies both original equations, we can confidently say that x = 5/13 and z = 1/13 is the correct answer. Checking your solution might seem like an extra step, but it's a valuable habit to develop. It can save you from submitting incorrect answers and gives you peace of mind knowing you've nailed the problem.
Alternative Methods
While we focused on the elimination method here, it's worth mentioning that there are other ways to solve systems of equations. The substitution method is another popular technique. In this method, you solve one equation for one variable and then substitute that expression into the other equation. This leaves you with a single equation in one variable, which you can solve. Then, you plug that value back into one of the original equations to find the other variable.
Another approach, especially for larger systems of equations, is to use matrices. Matrix methods, such as Gaussian elimination, provide a systematic way to solve systems with many variables and equations. These methods are often used in computer programs and are incredibly efficient.
Choosing the best method depends on the specific problem. For simple systems like the one we tackled today, elimination or substitution are often the quickest routes. But for more complex systems, matrices might be the way to go. The key is to understand the strengths and weaknesses of each method and choose the one that best fits the situation.
Real-World Applications
You might be wondering,
